难度: 困难(Hard)

题目

Follow up for “Find Minimum in Rotated Sorted Array”:
What if duplicates are allowed?

Would this affect the run-time complexity? How and why?

Suppose a sorted array is rotated at some pivot unknown to you beforehand.

(i.e., 0 1 2 4 5 6 7 might become 4 5 6 7 0 1 2).

Find the minimum element.

The array may contain duplicates.

代码

class Solution(object):

    def search(self, nums, l, r):
        if l==r:
            return nums[l]
        m = l + (r-l)/2
        if nums[m] == nums[l]:
            return min(self.search(nums,l,m), self.search(nums,m+1,r))
        elif nums[m] > nums[l]:
            return min(nums[l], self.search(nums, m+1, r))
        else:
            return self.search(nums, l, m)

    def findMin(self, nums):
        """
        :type nums: List[int]
        :rtype: int
        """
        if nums==None or len(nums)==0:
            return None
        return self.search(nums,0,len(nums)-1)